Chemistry(Please check)
posted by Hannah on .
What is molar solubility of Al(OH)3 at 25 degrees C given the solubility product Ksp = 5.0e33?
I set this up as Al^3+ + 3OH^
Ksp=[Al^3+][OH^]^3
5.0e33 = (x)(3x)^3
Before I go any further did I do this correctly?

you are right, I was wrong yesterday.
solve for x. 
So it would be 27x^4, so I divide 5.0e33 by 27 to get 1.8e34 and then raise this to the 1/4 power which I got 4e9.

I get 3.69E9